Subset feedback vertex set is fixed parameter tractable
Marek Cygan Marcin Pilipczuk Michał Pilipczuk Jakub Onufry Wojtaszczyk∗
Abstract
The FEEDBACKVERTEXSETproblem, given an undirected graphGand an integerk, asks whether there exists a set of at mostkvertices that hits all cycles in the graphG. FEEDBACKVERTEXSETwas intensively studied from the parameterized perspective. In this paper we consider the SUBSETFEED
-BACKVERTEXSETproblem, where an instance comes additionally with a subset of verticesSand we ask for a set of at mostkvertices that hits all simple cycles passing throughS. The parameterized com-plexity of SUBSETFEEDBACKVERTEXSET was posted as an open problem by S. Saurabh in 2009. We begin by showing that this problem is FPT when parameterized by|S|. Moreover, we present an algorithm which reduces the size of|S|toO(k3)using kernelization techniques such as the2-Expansion Lemma, Menger’s theorem and Gallai’s theorem. These two results prove that the SUBSETFEEDBACK
VERTEXSETproblem is FPT when parameterized by kand the time complexity of our algorithm is 2O(klogk)nO(1).
Furthermore, we present a connection between the SUBSETFEEDBACKVERTEXSETproblem and terminal separation problems such as MULTIWAYCUTand MULTICUT. We show that SUBSETFEED
-BACKVERTEXSETis a generalisation of MULTIWAY CUTand with an additional constraint SUBSET
FEEDBACK VERTEXSET becomes as hard as MULTICUT. It is known that MULTIWAY CUTis FPT when parameterized by the solution sizekbut for MULTICUTthe existence of such a FPT algorithm is a major and long-standing open problem in the FPT field. Our results imply that in some sense SUBSET -FVS falls between MULTIWAYCUTand MULTICUT. Since SUBSET-FVS is at least as hard as NODE
MULTIWAYCUTand it is still FPT, we make a step towards solving the MULTICUTproblem.
∗
Dept. of Mathematics, Computer Science and Mechanics, University of Warsaw, Poland,
1
Introduction
In the parameterized complexity setting, an instance comes with an integer parameter k — formally, a parameterized problemQis a subset ofΣ∗×Nfor some finite alphabetΣ. We say that a problem isfixed
parameter tractable(FPT) if there exists an algorithm solving any instance(x, k)in timef(k)poly(|x|)for some (usually exponential) computable functionf. Intuitively, the parameterk measures the hardness of the instance.
FEEDBACKVERTEXSET(FVS) is one of the long–studied problems in the parameterized complexity
setting. It can be stated as follows: given an undirected graphGonnvertices and a parameterkdecide if one can remove at mostkvertices fromGsuch that the remaining graph does not contain a cycle, i.e., is a forest. The long line of research concerning the FVS problem contains [2, 10, 11, 19, 20, 15, 16, 8, 6, 1]. Currently the fastest known algorithm works in 3.83knO(1) time [5]. Recently Thomass´e [23] showed a quadratic kernel for this problem improving previous results [4, 3]. The directed version has been proved to be FPT by Chen et al. [7].
Notation. Let us now introduce some notation. LetG = (V, E) be a simple undirected graph withn vertices. AcycleinGis a sequence of verticesv1v2. . . vm ∈ V such thatvivi+1 ∈E andvmv1 ∈E. We
say a cycle issimpleifm >2and the verticesviare pairwise different. We shall also consider multigraphs
(i.e., graphs with multiple edges and loops), in which a simple cycle can have two vertices if there is a multiple edge between them. We call an edgevw ∈ E abridgeif in(V, E\ {vw})the verticesvandw are in different connected components. Note that no simple cycle can contain a bridge as one of its edges. Given subsetsX, Y ⊆V, byE(X, Y)we denote the set of edges with one endpoint inX and the other in Y. ByG[X]we denote the subgraph induced byX with the edge setE(X, X). ByN(X)we denote the neighbourhood ofX, i.e.{u∈V :∃v∈Xuv ∈E} \X. For a subset of edgesE0 ⊆EbyV(E0)we denote
the set of all endpoints of edges from the setE0.
Problem definitions. In this paper we study the SUBSET FEEDBACK VERTEX SETproblem (SUBSET -FVS), where an instance comes with a subset of verticesS, and we ask for a set of at mostkvertices that hits all simple cycles passing throughS. It is easy to see that SUBSET-FVS is a generalisation of FVS by
puttingS =V.
SUBSETFEEDBACKVERTEXSET(SUBSET-FVS) Parameter:k
Input:An undirected graphG= (V, E), a setS ⊆V and a positive integerk
Question: Does there exist a setT such that|T| ≤ kand no simple cycle inG[V \T]contains any vertex ofS?
SUBSET-FVS was studied from the approximation perspective and the best known approximation
algo-rithm by Even at al. [12] gives approximation ratio equal to8. The parameterized complexity of SUBSET
FEEDBACKVERTEXSETwas posted as an open problem by S. Saurabh at Dagstuhl seminar 09511 [9]. In
this paper we answer this question by showing a2O(klogk)nO(1)algorithm.
Since we would like to use the properties of bridges, we define the following problem that we show to be equivalent to SUBSET-FVS.
EDGESUBSETFEEDBACKVERTEXSET(EDGE-SUBSET-FVS) Parameter:k
Input:An undirected graphG= (V, E), a setS ⊆Eand a positive integerk
Question: Does there exist a setT ⊆V with|T| ≤k, such that no simple cycle inG[V \T]contains an edge fromS?
(G, S0, k) of EDGE-SUBSET-FVS by selecting as S0 all the edges incident to any vertex of S. Then any
simple cycle passing through a vertex of S has to pass through an edge of S0, and conversely, any cycle passing through an edge ofS0 contains a vertex fromS. In the other direction, if(G, S0, k)is an instance
of EDGE-SUBSET-FVS, obtainG0by replacing each edgeuv ∈S0by a pathu−xuv−vof length2, and
solve the SUBSET-FVS instance (G0, S, k) whereS = {xe : e ∈ S0}. Clearly both reductions work in
polynomial time and do not change the parameter. Thus, in the rest of this paper we focus on solving EDGE
SUBSETFEEDBACKVERTEXSET. A simple cycle containing an edge fromS is called anS–cycle.
Our contributions. We develop a2O(klogk)nO(1) algorithm for EDGE-SUBSET-FVS (which implies an algorithm of the same time complexity for SUBSET-FVS). This result resolves an open problem posted by Saket Saurabh [9]. To achieve this result we use several tools such as iterative compression, the2-Expansion Lemma, Menger’s theorem, Gallai’s theorem and the algorithm for the MULTIWAYCUTproblem.
Moreover, we present a connection between the EDGESUBSETFEEDBACKVERTEXSETproblem and terminal separation problems such as MULTIWAY CUT and MULTICUT. We show that EDGE SUBSET
FEEDBACK VERTEX SETis a generalisation of NODE MULTIWAYCUT and with an additional constraint
EDGE-SUBSET-FVS becomes as hard as MULTICUT. This implies that in some sense EDGE-SUBSET
-FVS falls between MULTIWAYCUTand MULTICUT. However the parameterized status of the MULTICUT
problem is a major and long-standing open problem in the FPT field. Since EDGE-SUBSET-FVS is at least as hard as NODEMULTIWAYCUTand it is still FPT, we make a step towards solving the MULTICUT
problem. As EDGE-SUBSET-FVS is equivalent to SUBSET-FVS, these results hold for the latter problem as well.
Some of our ideas were inspired by the previous FPT results: the algorithm for MULTICUT parame-terized by(|T|, k) by Guillemot [14], the37.4knO(1)–time algorithm for FVS by Guo et al. [15] and the quadratic kernel for FVS by Thomass´e [23].
Outline of the paper. In Section 2 we show a relation between the variants of the FVS problems and terminal separation problems. In Section 3 we present an FPT algorithm for EDGE-SUBSET-FVS when parameterized by|S|, where for the sake of presentation we give an easy-to-describef(|S|)nO(1)algorithm at the cost of a fast growing functionf. This algorithm is enhanced in Appendix B so that the functionf becomes2O(klog|S|). Later in Section 4 we develop a set of reductions which work inf(k)nO(1) time and reduce the size of the setStoO(k3). The detailed proof of a crucial lemma from Section 4 is postponed to Appendix A due to space limitations and in order to make Section 4 easier to follow. We give an extended summary of the last step of Section 4 in Appendix C.
2
E
DGE-S
UBSET-FVS
related to terminal separation
In order to show the importance of our result for the EDGESUBSETFEEDBACKVERTEXSETproblem, we define two well-known terminal separation problems.
NODEMULTIWAYCUT(EDGEMULTIWAYCUT) Parameter:k
Input:An undirected graphG= (V, E), a set of terminalsT⊆V, and a positive integerk
Question: Does there exist a setT ⊆V (T ⊆E) such that|T| ≤kand no pair of terminals from the setT is contained in one connected component of the graphG[V \T](G0= (V, E\T))?
NODEMULTICUT(EDGEMULTICUT) Parameter:k Input: An undirected graphG = (V, E), a set of pairs of terminalsT = {(s1, t1), . . . ,(sm, tm)}, and a
positive integerk
Question: Does there exist a setT ⊆V (T ⊆E) such that|T| ≤kand no pair of terminals from the setT is contained in one connected component of the graphG[V \T](G0= (V, E\T))?
Both the vertex and edge versions of the MULTIWAYCUTproblem are known to be FPT since 2004 [17]. However the status of both versions of the MULTICUT problems is unknown and remains a major long-standing open problem in the FPT field (see [9]). It is known (e.g. [12]) that in the weighted case the NODE
MULTIWAY CUT problem can be reduced to weighted SUBSET-FVS by adding a vertex s(with infinite
weight) to the graph and connecting it to all the terminals, whereS = {s}. However, in the unweighted case this reduction does not hold. Hence we present a modified version.
Theorem 1. An instance(G,T, k)of theNODEMULTIWAYCUTproblem can be transformed in polynomial time into an equivalent instance(G0, S, k)of theEDGESUBSETFEEDBACKVERTEXSETproblem.
Proof. LetT = {v1, . . . , vt}. We add a setT0 = v01, . . . , vt0 oftvertices to the graph Gobtaining a new
graph G0. Together with the vertices from the set T0 we add a set of edges S = {vivi0 : 1 ≤ i ≤ t}.
Moreover, we add an edge between every pair of vertices from the setT0 so that G[T0]becomes a clique. Assume that(G, S, k)is a YES-instance of NODEMULTIWAYCUTwhereT ⊆V is a solution. ClearlyT is a solution for the instance(G0, S, k)of EDGE-SUBSET-FVS since anS-cycle inG0[(V ∪T0)\T]implies a path between terminals inG[V \T]. In the other direction, assume that (G0, S, k)is a YES-instance of
EDGE-SUBSET-FVS whereT ⊆V is a set of removed vertices. Observe that sinceG0[T0]is a clique there
is no point in removing any vertexv0isince we can removeviinstead. Hence we may assume thatT∩T0 =∅
which means thatT is a solution for the instance(G,T, k)of NODEMULTIWAYCUT.
On the other hand, let us consider an FVS-type problem where we are given a subset of edgesSand we are interested in hitting all simple cycles that pass through exactly one edge from the setS.
SPECIALEDGESUBSETFEEDBACKVERTEXSET(SPECIAL-EDGE-SUBSET-FVS) Parameter:k
Input:An undirected graphG= (V, E), a setS ⊂Eand a positive integerk
Question:Does there exist a setT such that|T| ≤kand no simple cycle inG[V \T]contains exactly one edge from the setS?
Observe that in SPECIAL-EDGE-SUBSET-FVS edges from the setSdo not have to be bridges inG[V \
T]because two edges from the setScan possibly lie on one cycle. We show that SPECIAL-EDGE-SUBSET -FVS is equivalent to the NODEMULTICUTproblem.
Theorem 2. An instance(G,T, k)of theNODEMULTICUTproblem can be transformed in polynomial time into an equivalent instance(G0, S, k) of theSPECIAL EDGE SUBSET FEEDBACKVERTEX SETproblem and, conversely, an instance ofSPECIALEDGESUBSETFEEDBACKVERTEXSETcan be transformed into an instance ofNODEMULTICUT.
Proof. Consider an instance(G,T, k)of NODEMULTICUT. For each pair of terminals(si, ti)∈Twe add
a vertexuiand edgessiuianduitito the graphGobtaining a graphG0. We takeSto be the set of the edges
siui. Observe that each simple cycle passing through exactly one edge fromSin the graphG0 corresponds
to a path between some pair of terminals from the setTin the graphGand vice-versa.
In the other direction we consider an instance(G0, S, k)of SPECIALEDGESUBSETFEEDBACKVER -TEX SET. We construct a setTwhich contains a pair(u, v)for each edgeuv ∈ S. Now we ask for NODE
3
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DGE-S
UBSET-FVS
parameterized by
|
S
|
In this section we concentrate on solving the EDGE-SUBSET-FVS problem parameterized by|S|, which means that our complexity function can be exponentially dependent on the number of edges in the setS. This is the first step towards obtaining a FPT algorithm when parameterized byk. Observe that we may assumek < |S|since otherwise we may delete one vertex from each edge from the setS thus removing all edges from the setSfrom our graph. Firstly, we give a FPT algorithm which is easy to understand and later we present methods to improve the time complexity. In this section we use the fact that NODEMULTICUT
is FPT when parameterized by(k,|T|)which was shown by Marx [18].
Theorem 3. There exists an algorithm solving theEDGE SUBSET FEEDBACK VERTEX SET problem in
f(|S|)nO(1)time, for some functionf.
Proof. LetTbe some solution of EDGE-SUBSET-FVS. Our new parametrization, by|S|, allows us to guess,
by checking all possibilities, the subsetTS =T ∩V(S)that is removed by the solutionT. Moreover, our
algorithm guesses how the setV(S)\TS is partitioned into connected components in the graphG[V \T]
with the edges fromSremoved. Clearly both the number of subsets and of partitions is a function of|S|. For a partitionP={P1, . . . , Pm}ofV(S)\TS we form a multigraphGPon the set{P1, . . . , Pm}by adding
an edgePiPjfor every edgeuv∈S, whereu∈Pi, v ∈Pj. Now we check whether there exists an edge in
GPwhich is not a bridge. If that is the case we know that the partitionPdoes not correspond to any solution
of EDGE-SUBSET-FVS, as any simple cycle inGPcan be converted into a simple cycle inG, hence we skip
this partition. Otherwise we create a set of pairsT, containing all pairs of vertices from the setV(S)\TS
that belong to different sets in the partitionP. FormallyT={(vi, vj) :vi ∈Pi0, vj ∈Pj0, i0 6=j0}. Because
of the properties of the multigraphGPit is sufficient to ensure that no pair from the setTis contained in one connected component, hence the last step is calling an algorithm for the NODE MULTICUTproblem with parameterk− |TS|. Note that even if its answerXis not disjoint withV(S)\TS,TS∪Xis still a solution to
EDGE-SUBSET-FVS: the connected components ofG[V \(TS∪X)]induce a partition ofV(S)\(TS∪X)
that is a subpartition ofPand thus all remaining edges ofS are bridges inG[V \(TS∪X)]. On the other
hand, if the answer to EDGE-SUBSET-FVS is positive, solution can be found in this manner for at least one choice ofTS andP. Observe that|T| = O(|S|2)so we obtain an FPT algorithm for the EDGE SUBSET
FEEDBACKVERTEXSETproblem parameterized by|S|.
Function EdgeSubsetFeedbackVertexSet(G, S, k){parameterized by(|S|, k)}
1: for allsubsetsTS ⊆V(S),|TS| ≤kdo
2: for allpartitionsP={P1, . . . , Pm}ofV(S)\TS do
3: form a multigraphGP on the set{P1, . . . , Pm}by adding an edgePiPj for every edgeuv ∈ S,u ∈ Pi,
v∈Pj.
4: ifall edges inGPare bridgesthen
5: letT={(vi, vj) :vi ∈Pi0, vj ∈Pj0, i06=j0}
6: ifMultiCut(G[V \TS],T, k− |TS|)returns(Y ES, X) then
7: returnTS∪X
8: returnN O
Using techniques developed by Guillemot [14] we can obtain a better time complexity. The detailed proof of the following theorem can be found in Appendix B.
Theorem 4. There exists an algorithm solving theEDGE SUBSET FEEDBACK VERTEX SET problem in 2O(klog|S|)nO(1)time.
4
E
DGE-S
UBSET-FVS
parameterized by
k
In this section we show an FPT algorithm for EDGE SUBSET FEEDBACK VERTEX SET that works in
2O(klogk)nO(1)time.
We follow the idea ofiterative compressionproposed by Reed et al. [21]. Firstly, note that ifV0 ⊆ V andZis a feasible solution to an EDGE-SUBSET-FVS instance(G, S, k), thenV0∩Zis a feasible solution to the EDGE-SUBSET-FVS instance(G[V0], S0, k), whereS0 = S ∩E(G[V0]). Thus, if the answer for
(G[V0], S0, k)is negative, so is the answer for(G, S, k). LetV ={v1, v2, . . . , vn}be an arbitrary ordering
of the set of vertices of G. We consecutively construct solutions to EDGE-SUBSET-FVS for instances
(G[Vi], Si, k), where Vi = {v1, v2, . . . , vi} and Si = S ∩E(G[Vi]). When looking for a solution for
graphG[Vi+1], we use the fact that ifZi is a solution for(G[Vi], Si, k), thenZ∪ {vi+1}is a solution for
(G[Vi+1], Si+1, k+ 1) — a solution for our problem with the parameter increased by one. Thus, when
solving the EDGE-SUBSET-FVS instance(G, S, k)we may access the setZ — a feasible solution for the instance(G, S, k+ 1).
Our strategy is to provide a set ofreductionsthat leads to an equivalent instance of EDGE-SUBSET-FVS
with|S|=O(k3)and then apply the algorithm described in Section 3. Denote byk0the initial value of the
parameterk— the value of the parameter may drop during reductions. We assumek≤k0and|Z| ≤k0+ 1.
More formally, every ourreductionis a polynomial–time algorithm that satisfies the following conditions: 1. as an input it takes an instance(G, S, k),k≤k0, together with a set of verticesZthat is a solution to
the instance(G, S,|Z|)with|Z| ≤k0+ 1;
2. returns either an instance(G0, S0, k), whereG0 has a smaller total number of edges and vertices than G, or at mostk0+ 1instances(Gi, Si, ki), for which allkis are strictly smaller thank;
3. in the first case (G, S, k) is a YES–instance iff (G0, S0, k) is a YES–instance, in the second case
(G, S, k)is a YES–instance iff there exists an indexisuch that(Gi, Si, ki)is a YES–instance;
4. there exists a polynomial–time algorithm that, given a feasible solution for (Gi, Si, ki) produces a
feasible solution for (G, S, k); in our reductions this algorithm is always trivial and we skip it in further descriptions.
In other words, in each reduction we either reduce the size of the graphG, or we branch into at mostk0+ 1
options, while in each option we strictly decrease the parameter. If we branch, we solve all the new instances recursively.
For each instance that is not applicable to any of our reductions, we claim that then |S| = O(k03)and pass this instance to the algorithm described in Section 3. As at each step we branch into at mostk0 + 1
options, the depth of the search tree is at mostk0, and the size of the graph can be decreased onlynO(1)
times, we obtain an algorithm working in time2O(klogk)nO(1).
In next subsections we describe a number of reductions. We assume that at each step, the lowest– numbered applicable reduction is used. If no reduction is applicable, we run the algorithm of Section 3 on the instance.
4.1 Simple reductions
First, we develop a few simple reductions. During other reductions, our graph may gain some multiple edges or loops. The following trivial reductions sort them out and remove clearly insignificant parts ofG.
Reduction 1. If there is a loopvvnot belonging toS, remove it. If there is a loopvvbelonging toS, remove vand decreasekby one —vneeds to be in any solution to(G, S, k).
Reduction 2. Assumeuandvare connected by multiple edges. If at least one of the edges belongs toS,u orvneeds to be in any feasible solution: branch, either removinguorv, in both cases decreasekby one. If none of the edges belong toS, remove all but one of these edges.
Reduction 3. Remove all bridges and connected components not containing any edge fromS.
Lemma 5. If none of the Reductions 1–3 is applicable, the graph G is simple and every edge in S in contained in some simple cycle.
4.2 The outer–abundant lemma
In this section we consider an instance of EDGE-SUBSET-FVS(G, S, k), whereG = (V, E). We assume that Reductions 1–3 are not applicable, i.e.,Gis simple and every edge inS belongs to some simple cycle. The approach here is based on ideas from the quadratic kernel for the classical FEEDBACK VERTEX SET
problem [23], however, a few aspects need to be adjusted to better fit our needs. Definition 6. A setF ⊆V is calledouter–abundantiff:
(a) G[F]is connected,
(b) there are no edges fromS inG[F],
(c) there at least10kedges fromS incident withF.
Lemma 7(The outer–abundant lemma). LetF be an outer–abundant set. If reductions 1–3 are not appli-cable, then in polynomial time one can find a nonempty setX ⊆V \F such that the following condition is satisfied: if there exists a solutionAforEDGE-SUBSET-FVSon(G, S, k)such thatA∩F =∅, then there exists a solutionA0 such thatA0∩F =∅andX ⊆A0.
The proof of Lemma 7 can be found in Appendix A and it involves kernelization techniques such as the
2-Expansion Lemma, Menger’s theorem and Gallai’s theorem.
As a direct application of Lemma 7, we obtain the following reduction rule. Note that if it is not applicable, there are at most10k|Z|=O(k20)edges fromSincident withZ.
Reduction 4. Letvbe a vertex that is incident to at least10kedges fromS. Apply Lemma 7 to the outer– abundant setF = {v}obtaining a setX. Then, ifvis not a part of a solution, we may greedily takeX into the solution. Therefore we branch: either we removev and decreasekby one, or we removeX and decreasekby|X|.
4.3 Bubbles
As discussed in the introduction to this section, we need to complete our reduction rules so that, givenZ — a solution to(G, S,|Z|), we decrease the size ofStoO(k03). After Reduction 4, there areO(k20)edges from Sincident withZ. Thus, we need to care only aboutS∩E(G[V \Z]).
AsZ is a feasible solution to EDGE-SUBSET-FVS on(G, S,|Z|), every edge fromS∩E(G[V \Z])
has to be a bridge inG[V \Z]. After removing those bridgesG[V \Z]becomes an union of connected components not having any edge from S. We call each such a component a bubble. Denote the set of
bubbles byD. OnDwe have a natural structure of a graphH = (D, ED), whereIJ∈EDiff components I andJ are connected by an edge fromS. AsZ is a solution,His a forest and eachI, J connected inH are connected inGby a single edge fromS.
Consider I ∈ D. Denote the set of vertices of I by VI. Note that if at most a single edge leaves I
(that is, |E(VI, V \VI)| ≤ 1) thenVI would be removed while processing Reduction 3. The following
reduction sorts out bubbles with exactly two outgoing edges and later we assume that for everyI ∈ Dwe have|E(VI, V \VI)| ≥3.
Reduction 5. Let us assume that|E(VI, V\VI)|= 2and let{u, v}=N(VI)(possiblyu=v). Each cycle
passing through a vertex fromVIis either fully contained inIand thus non–S–cycle, or exitsVI throughu
andv. We removeVIfrom the graph and replace it with a single edgeuv(possibly creating loop or multiple
edge), belonging toSiff any one of the two edges inE(VI, V \VI)is inS.
We are now left with bubbles that have at least three outgoing edges. We classify those bubbles according to the number of edges that connect them to other bubbles, that isdegH(I).
Definition 8. We say that a bubbleI ∈Dis (a) a solitary bubbleifdegH(I) = 0, (b) a leaf bubbleifdegH(I) = 1, (c) an edge bubbleifdegH(I) = 2, (d) an inner bubbleifdegH(I)≥3.
Denote byDs,Dl,De,Dithe sets of appropriate types of bubbles.
We show that we can do some reductions to make following inequalities hold:
|Dl|=O(k30), |Di|<|Dl|, |De|<3(|Z|+k) +|Di|+|Dl|.
Note that|Z| ≤ k0 + 1, so these conditions imply that|D\Ds|= O(k03). As edges ofH create a forest
overD\Ds, this bounds the number of edges fromSnot incident withZ byO(k03), as desired. Lemma 9. |Di|<|Dl|.
Proof. AsH is a forest, then|ED|< |D|= |Di|+|De|+|Dl|. Moreover,2|ED|= PI∈DdegH(I) ≥ 3|Di|+ 2|De|+|Dl|. Therefore|Dl|>|Di|.
Reduction 6. If |De| ≥ 3(|Z|+k) + |Di|+|Dl|, then we branch out into |Z| branches. In each of
the branches we remove a single vertex fromZ and decreasekby one (thus checking for solutions which contain this vertex).
For this reduction to be correct we have to prove that under the assumptions above every feasible solution of(G, S, k)contains a vertex fromZ — thus if each of our branches is a NO–instance, then(G, S, k)itself is a NO–instance.
Proof. We prove the correctness of the reduction by contradiction. As edge bubbles have degree2inH, H[De]is a set of paths of non-zero length and isolated vertices. Assume there are at least3(|Z|+k)vertices
contained in the paths. LetM be a maximal matching inH[De]. If a path containslvertices (forl ≥2),
it has a matching of cardinalitybl
2c ≥
l
3. Therefore, inH[De]we have a matching of cardinality at least
|Z|+k. Let us examine an arbitraryIJ ∈M. Recall that at least three edges leave each bubble. As each of VI,VJ is adjacent to two other bubbles by single edges, it has to be connected toZ as well. ChooseuI, uJ
— vertices fromZ such that uI ∈ N(VI) anduJ ∈ N(VJ). We see that there is a path from uI touJ
passing through an edge fromS: it goes fromuItoI, then toJ through an edge fromS, and then touJ. As
M is a matching, such paths are vertex–disjoint for allIJ ∈M, except for endpointsuIanduJ.
If we have a solution disjoint withZof cardinality at mostk, there are at least|Z|pairsIJ ∈M, where neither I nor J contains a vertex from the solution. Now we construct a graph P = (Z, EP) such that
uIuJ ∈ EP ifuI, uJ have been chosen for some IJ ∈ M, where I and J are solution–free. We prove
thatP has to be a forest. Indeed, otherwise there would be a cycle in P — and by replacing each edge from it by associated path, we construct anS-cycle inG(as paths in which edges fromEP originated are
vertex–disjoint). This cycle does not contain any vertex from the solution, as it passes only throughZ and solution–free bubbles. However, as|EP| ≥ |Z|,P cannot be a forest. Contradiction, so there are less than 3(|Z|+k)vertices contained in paths of nonzero length.
To end the proof we show that the number of edge bubbles not adjacent to any other edge bubble inH is at most|Di|+|Dl|. Let us root each connected component ofH in an arbitrary leaf and for any bubble
I ∈ De letϕ(I)be the only child of I. Observe that this mapping is injective and maps the set of edge bubbles isolated inH[De]intoDi∪Dl.
We are left with the leaf bubbles and we need to show reductions that lead to|Dl|=O(k03). We do this by a single large reduction encapsulated in the following lemma:
Lemma 10. There exists a proper reduction, which takes an instance(G, S, k)such that none of the Reduc-tions 1 – 6 is applicable, and either branches out into at mostk0+ 1instances(Gi, Si, ki)withki < kor returns a single equivalent instance with|S|=O(k03).
4.4 The leaf bubble reduction
We describe the final reduction in this subsection. It proceeds in a number of steps. Each step either branches out (thus ending the reduction) or — after, possibly, modifyingG— passes to the next step.
LetI be a leaf bubble. As there are at least three edges leavingI, each leaf bubble is connected toZ by at least two edges. We begin with a bit of preprocessing:
Step 1. As long as there are two verticesv, v0inZ withvv0 ∈/ E, and a set{I1, I2, . . . , Ik+1}of bubbles,
where eachIj is connected to bothvandv0 by edges not inS, we add an edgevv0toE, withvv0 ∈/ S. Lemma 11. The output(G0, S, k)of Step 1 is a YES–instance iff the input(G, S, k)is a YES–instance. Proof. Obviously if a solution exists to(G0, S, k), then it is a solution to(G, S, k), as we only added edges (and thus only added potential S–cycles). On the other hand, suppose we have a solutionT to(G, S, k)
which is not a solution to(G0, S, k). Then there is someS–cycleC in(G0, S, k)not passing through any vertex ofT.Chas to pass through the edgevv0 (otherwise it would also be anS–cycle in(G, S, k)).
As|T| ≤k, there is at least one bubbleIj which is disjoint withT. Thus we can find a simple pathP
connectingvandv0, the interior vertices of which are all inIj. Note that asIj is a bubble and the edges tov
Gwhich is formed by replacing the edgevv0inCby the pathP0. AsCwas anS–cycle, there is some edge e ∈ S∩C. This edge is visited byC0 exactly once — as we took outvv0 ∈/ S and added edges fromP, which is disjoint withS. If we consider the graph spanned by edges fromC0, the edgeeis not a bridge in this graph, as the endpoints ofeare connected byC0\ {e}. Therefore,elies on some simple cycle contained inC0and the output instance of Step 1 is indeed equivalent to the input instance.
Now for each bubble we choose arbitrarily two of the edges connecting it toZ:eIande0I. Additionally
assume that ifS∩E(VI, Z) 6=∅theneI ∈S. LetvI andvI0 be the endpoints ofeI ande0I inZ (possibly
vI =v0I). We say that a bubbleI isassociatedwith verticesvI, vI0. Lemma 12. If there are at least|Z|2(k+ 2)leaf bubblesI such thate
I∈Sthen every feasible solution of (G, S, k)contains a vertex fromZ.
Proof. By the Pigeonhole Principle, there existv, v0 ∈ Z associated with at leastk+ 2of the considered leaf bubbles. Ifv=v0, there arek+ 2S–cycles sharing onlyv, each constructed from a different bubbleI by closing a path contained inI with edgeseIande0I. Therefore,vneeds to be part of any feasible solution.
Ifv 6= v0, one can similarly choose a path betweenv andv0 which contains an edge fromS through each bubbleI. Thesek+ 2paths are vertex–disjoint apart fromvandv0, so any feasible solution disjoint withZ leaves at least two of them solution–free. These two paths can be arranged into a solution–freeS–cycle, so any feasible solution is not disjoint withZ, as it containsvorv0.
Lemma 13. If there are at least|Z|2(k+ 1)leaf bubblesI such thatv
Iv0I∈S, then every feasible solution of(G, S, k)contains a vertex fromZ.
Proof. As before, there existv, v0 ∈Zassociated with at leastk+ 1considered leaf bubbles. These bubbles generate at leastk+ 1S–cycles, which are vertex–disjoint apart fromv, v0. Therefore, any feasible solution needs to includevorv0.
Due to Lemmata 12 and 13 the following step is proper: Step 2. If there are at least|Z|2(k+ 2)leaf bubblesIsuch thate
I ∈Sor at least|Z|2(k+ 1)leaf bubblesI
such thatvIvI0 ∈S, branch out into|Z|new instances. In each of these instances remove one of the vertices
ofZfrom the graph and decreasekby1.
Now suppose that two leaf bubblesI1, I2 are connected inH(form aK2inH), by an edgeeI1I2 ∈S.
Denote such a pair abubble–bar.
Lemma 14. If there are at least|Z|2(k+ 2)bubble–bars, then any feasible solution to(G, S, k)contains a
vertex fromZ.
Proof. By the Pigeonhole Principle, there existv, v0 ∈ Z such that there exist at least k+ 2bubble–bars
(I1, I2)withvI1 =vandvI2 =v
0. Ifv=v0, there arek+2S–cycles having onlyvin common (one through each bubble–bar), so any feasible solution has to containv. Ifv6=v0, there arek+ 2paths connectingvand v0and sharing onlyvandv0. Any solution disjoint withZ would leave at least two of them solution–free. Then these two paths could be arranged into a solution–freeS–cycle.
Step 3. If there are at least |Z|2(k+ 2)bubble–bars, branch out into|Z|new instances. In each of these
instances remove one vertex fromZfrom the graph and decreasekby one.
Summing all the obtained bounds, we can count almost all the leaf bubbles (possibly more than once) and bound their number byO(k03). The ones that are left satisfy the following definition:
Definition 15. A leaf bubbleI satisfying the following three conditions is called aclique bubble: (a) G[N(VI)∩Z]is a clique not containing any edge fromS,
(b) Iis connected toZby edges not belonging toS, (c) Iis connected to a non–leaf bubble.
Denote the only edge from S connecting a given clique bubbleI with its neighbour bubble bywIw0I,
withwI ∈VI.
Lemma 16. If there exists a feasible solutionT for(G, S, k), then there exists a feasible solutionT0, which is disjoint from all clique bubbles inG.
Proof. Let I be a clique bubble. Assume we have a feasible solution T, withT ∩VI 6= ∅. We show
T0 = (T \VI)∪ {w0I}is also a feasible solution. Consider anyS–cycleC inG[V \T0]. This cycle has to
pass throughVI(possibly multiple times), or it would be anS–cycle inG[V\T], contrary to the assumption
T was a feasible solution. Note that C has to enter and exit VI throughN(VI)∩Z, as the only vertex
inN(VI)\Z isw0I, which is removed byT0. But thenC can be shortened toC0 by replacing every part
contained inVI by a single edge inZ (as N(VI)∩Z is a clique). NowC0 is disjoint fromVI and is an
S–cycle due to the definition of the clique bubble. SoC0 is anS–cycle inG[V \T], a contradiction. Note that the only vertex we added toT wasw0I, which does not belong to a clique bubble (it does not belong even to a leaf bubble, from property (c) in the definition of clique bubbles). Thus we can apply this procedure inductively, at each step reducing the number of vertices inT contained in clique bubbles, until none are left.
Assume there is a vertexv ∈Z such thatv∈N(VIj)for some distinct clique–bubblesI1, I2, . . . , I10k. We show that the setF = {v} ∪S10k
j=1VIj is outer–abundant inG. Indeed, it is connected and due to the definition of bubbles and properties (b) and (c) of the clique bubble definition, the subgraphG[F]does not contain edges fromS. Moreover, there are at least10kedges fromS incident withG[F]— these are the edges connecting bubblesIjwith other bubbles, not contained inFas they are non–leaf ones due to property
(c). This enables us to formulate the key step:
Step 4. If there is a vertexv∈Zwhich is adjacent to at least10kclique bubbles, we apply Lemma 7 to the setF = {v} ∪S10k
j=1VIj to obtain a setX. We branch out into two instances — in the first we removev from the graph and reducekby one, in the second we removeXfrom the graph and reducekby|X|.
Suppose there is a feasible solutionT to(G, S, k). Due to Lemma 16 we may assumeT to be disjoint withF \ {v}. Thus eitherT containsv, or it is disjoint withF, and by Lemma 7 there exists a solution containingX. This justifies the correctness of Step 4.
Thus we managed to reach the state when the number of leaf bubbles is bounded by O(k03). As we modified only subgraphG[Z], the setsDi,De,Dlremain the same after modifications and we obtain graph
with|S|=O(k03). This completes the description of the2O(klogk)nO(1)algorithm for EDGE-SUBSET-FVS. We include a detailed summary of the leaf bubble reduction in Appendix C.
References
[1] Ann Becker, Reuven Bar-Yehuda, and Dan Geiger. Randomized algorithms for the loop cutset prob-lem. J. Artif. Intell. Res. (JAIR), 12:219–234, 2000.
[2] Hans L. Bodlaender. On disjoint cycles. Int. J. Found. Comput. Sci., 5(1):59–68, 1994.
[3] Hans L. Bodlaender and Thomas C. van Dijk. A cubic kernel for feedback vertex set and loop cutset.
Theory Comput. Syst., 46(3):566–597, 2010.
[4] Kevin Burrage, Vladimir Estivill-Castro, Michael R. Fellows, Michael A. Langston, Shev Mac, and Frances A. Rosamond. The undirected feedback vertex set problem has a poly() kernel. In IWPEC, pages 192–202, 2006.
[5] Yixin Cao, Jianer Chen, and Yang Liu. On feedback vertex set new measure and new structures. In
Proc. of SWAT’10, pages 93–104, 2010.
[6] Jianer Chen, Fedor V. Fomin, Yang Liu, Songjian Lu, and Yngve Villanger. Improved algorithms for feedback vertex set problems. J. Comput. Syst. Sci., 74(7):1188–1198, 2008.
[7] Jianer Chen, Yang Liu, Songjian Lu, Barry O’Sullivan, and Igor Razgon. A fixed-parameter algorithm for the directed feedback vertex set problem. InProc. of STOC’08, pages 177–186, 2008.
[8] Frank K. H. A. Dehne, Michael R. Fellows, Michael A. Langston, Frances A. Rosamond, and Kim Stevens. An o(2o(k)n3) fpt algorithm for the undirected feedback vertex set problem. InCOCOON, pages 859–869, 2005.
[9] Eric D. Demaine, Mohammad Taghi Hajiaghayi, and D´aniel Marx. Open problems from dagstuhl seminar 09511, 2009.
[10] Rodney G. Downey and Michael R. Fellows. Fixed parameter tractability and completeness. In Com-plexity Theory: Current Research, pages 191–225, 1992.
[11] Rodney G. Downey and Michael R. Fellows. InParameterized Complexity, 1999.
[12] Guy Even, Joseph Naor, and Leonid Zosin. An 8-approximation algorithm for the subset feedback vertex set problem. SIAM J. Comput., 30(4):1231–1252, 2000.
[13] Tibor Gallai. Maximum–minimum s¨atze und verallgemeinerte faktorem von graphen. Acta. Math. Acad. Sci. Hungaricae, 2:131–173, 1961.
[14] Sylvain Guillemot. Fpt algorithms for path-transversals and cycle-transversals problems in graphs. In
Proc. of IWPEC’08, pages 129–140, 2008.
[15] Jiong Guo, Jens Gramm, Falk H¨uffner, Rolf Niedermeier, and Sebastian Wernicke. Compression-based fixed-parameter algorithms for feedback vertex set and edge bipartization. J. Comput. Syst. Sci., 72(8):1386–1396, 2006.
[16] Iyad A. Kanj, Michael J. Pelsmajer, and Marcus Schaefer. Parameterized algorithms for feedback vertex set. InIWPEC, pages 235–247, 2004.
[17] D´aniel Marx. Parameterized graph separation problems. InProc. of IWPEC’04, pages 71–82, 2004. [18] D´aniel Marx. Parameterized graph separation problems. Theor. Comput. Sci., 351(3):394–406, 2006. [19] Venkatesh Raman, Saket Saurabh, and C. R. Subramanian. Faster fixed parameter tractable algorithms
for undirected feedback vertex set. InISAAC, pages 241–248, 2002.
[20] Venkatesh Raman, Saket Saurabh, and C. R. Subramanian. Faster fixed parameter tractable algorithms for finding feedback vertex sets. ACM Transactions on Algorithms, 2(3):403–415, 2006.
[21] Bruce A. Reed, Kaleigh Smith, and Adrian Vetta. Finding odd cycle transversals. Oper. Res. Lett., 32(4):299–301, 2004.
[22] Alexander Schrijver. A short proof of mader’s sigma-paths theorem. J. Comb. Theory, Ser. B, 82(2):319–321, 2001.
[23] St´ephan Thomass´e. A quadratic kernel for feedback vertex set. InProc. of SODA’09, pages 115–119, 2009.
A
Proof of the outer–abundant lemma
Lemma 17(The outer–abundant lemma, which is Lemma 7 restated). LetF be an outer–abundant set. If the graphGis simple and every edge ofSis contained in a simple cycle, then in polynomial time one can find a nonempty setX ⊆V \F such that the following condition is satisfied: if there exists a solutionAfor
EDGE-SUBSET-FVSon(G, S, k)andA∩F =∅, then there exists a solutionA0such thatA0∩F =∅and
X⊆A0.
Before we start proving Lemma 7, let us recall a few tools used in the quadratic kernel for FEEDBACK
VERTEXSET[23]. First, we recall the result of Gallai on finding disjointA–paths.
Theorem 18(Gallai [13]). LetA be a subset of vertices of a graphG. A path is called anA–path if its endpoints are different vertices inA. If the maximum number of vertex disjointA–paths is strictly less than
k+ 1, there exists a set of verticesB0 ⊆V of size at most2kintersecting everyA–path.
Moreover, it follows from Schrijver’s proof of the Gallai’s theorem [22] that Theorem 18 can be algo-rithmized: in polynomial time we can find either(k+ 1)disjointA–paths or the setB0.
The other theorem we need is the2–Expansion Lemma:
Theorem 19(2–Expansion Lemma, Theorem 2.3 in [23]). LetHbe a nonempty bipartite graph on bipar-tition(X, Y)with|Y| ≥2|X|and such that every vertex ofY has at least one neighbour inX. Then there exists nonempty subsetsX0 ⊆X,Y0 ⊆Y such thatN(Y0)∩X =X0and one can assign to eachx∈X0
twoprivateneighboursy1x, y2x∈Y0(i.e., eachy∈Y0is assigned to at most onex∈X0). In addition, such pair of subsetsX0,Y0can be computed in polynomial time in the size ofH.
Proof of Lemma 7. AnS–cycle is calledimportantif it contains an edge fromS∩E(F, V \F). A set of important cycles{C1, C2, . . . , Ct}is called at–flower, if the sets of verticesCi\F are pairwise disjoint.
Note that if there exists a vertexv ∈ V \F such that|E({v}, F)| ≥ 2andE({v}, F)∩S 6= ∅, then one can takeX ={v}. Indeed, any solution disjoint withF has to includev, since by connectivity ofG[F]
there is an important cycle contained inF ∪ {v}passing throughvvia at least one edge fromS. Thus we can assume that each vertex fromV \F is connected toF by a number of edges not belonging toSor by a single edge fromS.
Now we prove that in polynomial time we can find one of the following structures: either a(k+ 1)– flower, or a setBof at most3kvertices belonging toV \F such that each important cycle passes through at least one of them (further called a3k–blocker).
Let C be the set of those important cycles, which contain exactly two edges between F and V \F (intuitively, visitingF only once). Note that due to connectedness ofG[F], a set is a 3k–blocker iff any cycle fromCpasses through at least one of its elements. ForC∈Clet us examine these two edges between F andV \F. AsCis important, one or two of them belong toS. We say that such a cycle is of type I iff exactly one of these edges belong toSand of type II otherwise.
Firstly, we sort out the type I cycles. We removeF from the graph and replace it with two verticessand t. Also we add edges incident withsandt— for everyvw ∈E(F, V \F)withv ∈ F, we add edgesw ifvw ∈S and edgewtotherwise. By a simple application of the vertex max–flow algorithm and Menger’s theorem one obtains either a vertex–disjoint set of paths betweensandtof cardinalityk+ 1, or a set of at mostkvertices such that each such a path passes through at least one of them. Returning to the original graph transforms each path betweensandtinto a type I cycle, so we have found either a(k+ 1)-flower or ak-blocker of type I cycles.
Now, we deal with the type II cycles. LetJ ⊆ V \F be the set of vertices that are connected toF by an edge fromS. We remove temporarilyF from the graph and apply Theorem 18 to the setJ. Note that a set ofk+ 1vertex–disjointJ–paths correspond to a(k+ 1)–flower, and the setB0 is a2k–blocker of type II cycles.
Using both of these methods we obtain either a(k+ 1)–flower or, by adding blockers, a3k–blocker. Note that both algorithms run in polynomial time.
The existence of a(k+ 1)–flower immediately shows that a solutionAdisjoint withF does not exists, as each cycle belonging to a flower has to include at least one vertex fromA. Thus we are left only with the case of a3k-blocker. Let us denote it byB.
Let us examineG[V \(F∪B)]. LetH = (VH, EH)be any of its connected components. Note thatH
is connected toF by a number of edges not belonging toS or by a single edge fromS. Indeed, otherwise, due to connectedness of H and ofG[F], there would be an important cycle contained inG[F ∪VH]not
blocked byB. Using observation from the second paragraph of this proof, we may assume that each vertex fromBis connected toFby at most one edge fromS. So there are at most3kedges fromSbetweenBand F. As there are at least10kedges fromSbetweenF andV \F, we have at least7kconnected components ofG[V \(F ∪B)]connected toFby a single edge fromS.
We call a componentH easyif there is anS–cycle fully contained inH. If the number of easy com-ponents is larger thank, there is more thankvertex–disjointS–cycles, soAdoes not exist. Thus we may assume that there are at least6knon–easy components connected toF by a single edge fromS. We call themtoughcomponents.
LetH = (VH, EH)be a tough component. Observe thatN(VH)∩B 6=∅. Indeed, otherwise the only
edge betweenVH andV \VH would be the edge fromSconnectingHwithF and thus a bridge sorted out
by Reduction 3.
LetTbe the set of tough components. We construct a bipartite graph(B∪T, Eexp)such thatvH ∈Eexp
iffv ∈N(VH). Note that due to observation in the previous paragraph,(B∪T, Eexp)satisfy assumptions
of Theorem 19, as|T| ≥6k= 2·3k≥2|B|. So we have nonempty setsX ⊆BandY ⊆T such that for everyv ∈ B there are twoprivatetough componentsHv,1, Hv,2 ∈ Y withv ∈ N(VHv,i)fori= 1,2and B∩S
H∈Y N(H) =X.
Letv∈X. Note that due to the connectedness ofHv,iandG[F], there is anS–cycleCvpassing through
v— it goes fromvtoHv,1, then toF through an edge fromS, then toHv,2 through an edge fromSand
back tov. Assume thatAis a solution to the EDGE-SUBSET-FVS on(G, S, k) andA∩F = ∅. We see that cyclesCv for v ∈ X form a |X|–flower, so there are at least|X|vertices inA∩(X∪SH∈Y VH).
On the other hand, eachS–cycle passing through any vertex fromX∪S
H∈Y VH passes through a vertex
fromX. Indeed, eachH ∈ Y is connected toV \VH with a single edge incident with F and a number
of edges incident withX. Hence each cycle passing through a vertex fromS
H∈Y VH is fully contained in
someH ∈Y or goes from someH ∈Y toX. As eachH ∈Y is non–easy, cycles not incident withXdo not contain any edge fromS.
These observations prove that if we constructA0fromAby replacingA∩(X∪S
H∈Y VH)withX,A0
will be still a solution to EDGE-SUBSET-FVS. Thus the setXsatisfies all the required conditions.
B
Improving time complexity in
E
DGE-S
UBSET-FVS
parameterized by
|
S
|
In this section we use a restricted version of the NODEMULTIWAYCUTproblem where we cannot remove terminals. Our whole approach in this subsection is closely based to the arguments of Guillemot [14] for
RESTRICTEDNODEMULTIWAYCUT Parameter:k Input:A graphG= (V, E), a setT⊆V and a positive integerk
Question:Is there a setT ⊆V\Twith|T| ≤ksuch that no two vertices from the setTbelong to the same connected component ofG[V \T]?
An algorithm with O(4knO(1))running time for RESTRICTEDNODEMULTIWAY CUTwas presented
by Guillemot [14]. Our main result in this section is the following:
Theorem 20(Theorem 4 restated.). There exists an algorithm solving theEDGESUBSETFEEDBACKVER
-TEXSETproblem in2O(klog|S|)nO(1)time.
For G = (V, E) denote GS = (V, E \S). By apartition of a set Z we mean such a family P = {P0, P1, . . . , Pm}, thatPis are pairwise disjoint and their union isZ. We say a partitionP0is asubpartition
ofPif every element ofP0is contained in some element ofP, in this case we callPasuperpartitionofP0.
Proof. The algorithm works in three phases. In the first two phases we aim to divide the set of all endpoints of edges fromSinto a family of subsets. Each such subset is contained in a single component ofGS[V\T],
whereT is some choice of thekvertices to remove fromV. In the third phase we check whether we can, in fact, achieve such a partition of the endpoints of edges fromSinto connected components by removing at mostkvertices, and whether such a partition implies that we removed all cycles passing throughSfromG.
InitializeP0 =∅. The first phase works as follows:
1. Select a spanning forestF ofGS, letU =V(S)be the set of endpoints of edges fromS;
2. Ifk= 0proceed directly to phase 2;
3. As long as there are isolated vertices not fromU or leaves not fromU inF, remove them fromF; 4. As long as there are vertices of degree 2 not fromU inF, remove them fromF, and connect the two
neighbours of the removed vertex with an edge inF;
5. Branch out — one branch passes the resultant forestF to the second phase. In other|F|branches we select a vertex fromF which we remove fromGS. We decreasekby one, add the vertex toP0and
repeat phase 1;
Note that after the first four steps of phase oneF has at most2|U|vertices (as all its vertices of degree at most2are fromU). Thus in the fifth step we have at most2|U|+ 1branches. As we can branch out into phase1at mostktimes due to steps 2 and 5, this assures we have at most(2|U|+ 1)kentries into phase two from phase one. The internal workings of the first phase are obviously polynomial–time.
The second phase is somewhat more complicated, and aims at arriving at a partition of setU, based on the forestF received from phase one. LetP0be the set of vertices fromU which were removed by phase
one. LetP ={P0, P1, . . . , Pm}be the partition ofU given by phase one — that is ifC1, C2, . . . , Cm are
connected components ofF, thenPi=Ci∩U. Note that thePis are non–empty (perhaps with the exception
ofP0), for if some connected component ofF contained no vertex fromU, then it would be removed from
F completely during the third step of phase one. Now we proceed as follows: 1. Select at mostkedges fromF;
3. For each partitionP00ofU being at the same time a subpartition ofPand a superpartition ofP0 start phase three.
We want to check how many times each application of phase two enters phase three. AsF has at most
2|U|edges, we have at most(2|U|)k choices in the first step. Now, consider a connected component Ci
ofF, from which we removedsi edges. There are exactlysi+ 1elements ofP0 which are subsets ofCi.
The number of ways to combine these elements into a partition ofCi∩U is the(si+ 1)–st Bell number
Bsi+1 ≤(si+ 1)
si+1. The product of these is
Y
i
Bsi+1≤
Y
i
(si+ 1)si+1 = exp
X
i
(si+ 1) log(si+ 1)
!
≤exp
1 +X i
si
!
log 1 +X i
si
!!
= (1 +k)1+k,
where the last inequality follows from the standard Jensen’s inequality corollaryf(P
ixi) ≤
P
if(xi)for
a convex functionf withf(0) = 0and non–negativexiapplied forf(x) = (1 +x) log(1 +x). Thus for
each execution of phase two, phase three is executed at most(2|U|)k(k+ 1)k+1times. Phase three works as follows:
1. Take the partition P00 ofU given by phase two, form a multigraph on the set {P1, P2, . . . , Pm} by
adding an edgePiPj for every edgeuv ∈ S, whereu ∈ Pi, v ∈ Pj. If any of these edges is not a
bridge, return a negative answer from this branch;
2. Otherwise create a graphG0 by adding toGa vertexwi for everyPi ∈P00,i≥1and adding edges
connectingwitouj for eachuj ∈Ci; letW ={wi}mi=1;
3. Apply RESTRICTED NODEMULTIWAY CUT to(G0S[(V ∪W)\P0], W, k− |P0|). If it returns a
negative answer, we return a negative answer from this branch, otherwise we return T ∪P0 as a
solution, whereT is the set returned by RESTRICTEDNODEMULTIWAYCUT.
The execution of this phase takes4knO(1)time. Thus, as the first phase branches out into at most(2|U|+ 1)kexecutions of phase two, phase two branches out into at most(2|U|)k(k+1)k+1instances of phase three, and phase three executes in4knO(1), the runtime of the whole algorithm is at most2O(klog|U|)nO(1)(we may assumek≤ |U|— if otherwise, removingU fromV gives a trivial positive solution).
Now we prove the correctness of this algorithm. First assume the algorithm returns a solution. It was then found by phase three. Consider the graph G0[(V ∪W)\(T ∪P0)]. AsT was returned by the
RESTRICTED NODEMULTIWAY CUT algorithm, we know|T| ≤ k− |P0|(and thus |T ∪P0| ≤ k). We
prove there are no simple cycles through edges ofS inG0[(V ∪W)\(T ∪P0)], which implies the same
for its subgraph G[V \(T ∪P0)]. As T was returned by RESTRICTED NODE MULTIWAY CUT, each
vertex wi is in a distinct connected component ofG0S[(V ∪W)\(T ∪P0)]. Consider a simple cycle in
G0[(V ∪W)\(T ∪P0)]passing through some edgee∈S. This means there is a path connecting the two
endpoints ofeinG0[(V ∪W)\(T ∪P0)]not passing throughe. We may contract connected components
containing thewis to single vertices, thus receiving a graph isomorphic to the graph considered in the first
step of phase three. The existence of the simple path, however, meansewas not a bridge in this graph, contrary to the assumption our algorithm returned T ∪P0 as a solution. Thus any solution found by our
On the other hand, assume there exists some solutionT of the EDGE-SUBSET-FVS problem. We show
how our algorithm arrives at a positive answer in this case. In the first phase, if any vertext ∈ T remains in the forest F finally constructed in this phase, we add this vertex to P0 and branch out to remove this
vertex. If no vertex fromT remains inF at some iteration of the first phase, we branch out to the second phase. In the second phase, we choose the edges into which the vertices fromT were contracted (if some were dropped due to being leaves or isolated vertices inF, we simply choose fewer edges). Now consider the partitionP00ofU given by the relation of being in the same connected component ofG[V \T]. It is a subpartition ofP, asPis simply the partition given byG[V \P0], andP0⊆T. It is also a superpartition of
P0, asP0 is given by removing the wholeT and all the edges which are not edges of the forestF. ThusP00 is one of the partition considered in the second phase, and thus enters the third phase. Now for this partition the edges ofSare bridges in the sense of the first step of phase three, as the vertices of the graph considered there correspond exactly to connected components ofG[V\T], which has no simple cycles through edges in S. Moreover,T is a solution for the RESTRICTEDNODEMULTIWAYCUTproblem by its definition. Thus
RESTRICTEDNODEMULTIWAYCUTreturns some positive answer (not necessarilyT) in this branch, and
thus the algorithm gives the correct answer.
C
Summary of the leaf bubble reduction
In this section we summarize steps made in Section 4.4, to show clearly that the number of leaf bubbles is bounded byO(k30).
Assume we passed through all the steps without branching. Denote the input graph byG, and the output byG0. The only modifications we applied toGwere adding edges inZ (in Step 1). Let us check that the output graph indeed hasO(k30)edges fromS:
1. The decomposition of V(G)\Z into bubbles is the same as the decomposition ofV(G0)\Z and bubbles that were inner or edge bubbles inGare, respectively, inner or edge bubbles inG0;
2. There are at most|Z|2(k+ 2)−1leaf bubbles connected toZby an edge fromS, as we passed over
Step 2;
3. There are at most|Z|2(k+ 1)−1leaf bubbles associated with a pair of vertices connected with an
edge fromS, as we passed over Step 2;
4. There are at most2|Z|2(k+ 2)leaf bubbles connected to other leaf bubbles, as we passed over Step
3;
5. For any pair v, v0 of vertices in Z withvv0 ∈/ E there are at most kleaf bubbles adjacent to both vertices of that pair through edges not inS, for otherwise we would have added that edge in Step 1. 6. If a leaf bubble is not a clique bubble, it either is connected to a leaf bubble (forming a bubble–bar), is
connected toZby an edge inS, has an edge fromSbetween some two of its neighbours inZ, or has some two neighbours inZ not connected by an edge. The number of such bubbles in all four cases was estimated above.
7. Thus, in total, there are at most O(k30) bubbles which are not clique bubbles, and there are at most
8. Thus |Dl| = O(k03), moreover|Di| ≤ |Dl|by Lemma 9 and |De| ≤ 3(|Z|+k) +|Di|+|Dl|by Reduction 6 — thus the number of edges inS not incident withZ is bounded byO(k3
0). We added
no new edges toS, and the number of edges inS incident toZ was bounded byO(k02)in the input graph, thus in the output graph there areO(k30)edges fromS, as desired.
